A rectangular swimming pool $10 \, {\rm m}$ long and $5 \,{\rm m}$ wide has a depth of $3 \,{\rm m}$ at one end and $1 \,{\rm m}$ at the other end. If water is pumped into the pool at the rate of $300$ liters per minute (${\rm liter}/{\rm min}$), at what rate is the water level rising when it is $1.5 \,{\rm m}$ deep at the deep end?

Solution

Step 1 We draw a picture of the cross-sectional view of the pool, as shown in Figure 4.

Step 2 The width of the pool is $5 \,{\rm m}$, the water level (measured at the deep end) is $h$, the distance from the wall at the deep end to the edge of the water is $L$, and the volume of water in the pool is $V$. Each of the variables $h$, $L$, and $V$ varies with respect to time $t$. $${\rm We\;are\;given} \dfrac{\textit{dV}}{\textit{dt}}=300\, {\rm liter}/ {\rm min}\; {\rm and\;are\;asked\;to\;find} \dfrac{\textit{dh}}{\textit{dt}} {\rm when}\; h=1.5\; {\rm m}.$$

Step 3 The volume $V$ is related to $L$ and $h$ by the formula $$V=\hbox{(Cross-sectional triangular area)(width)}=\left( \dfrac{1}{2} Lh\right) (5) =\dfrac{5}{2}Lh$$

See Figure 4. Using similar triangles, $L$ and $h$ are related by the equation $$\frac{L}{h}=\frac{10}{2} \qquad\hbox{so} \qquad L=5h$$

Now we can write $V$ as $$\begin{eqnarray*} V = \frac{5}{2} \textit{Lh} \underset{\underset{{\color{#0066A7}{\hbox{L=5h}}}}{\color{#0066A7}{\uparrow }}} {=} \frac{5}{2}(5h) h=\dfrac{25}{2}h^{2} \nonumber\\[-8.1pt]\tag{1} \end{eqnarray*}$$

Both $V$ and $h$ vary with time $t$.

258

Step 4 We differentiate both sides of equation (1) with respect to $t.$ $$\dfrac{\textit{dV}}{\textit{dt}}=25h\,\frac{\textit{dh}}{\textit{dt}}$$

Step 5 Substitute $h=1.5$ m and $\dfrac{\textit{dV}}{\textit{dt}}=300\ {\rm liter}/{\rm min}= \dfrac{300}{1000}{\rm m}^{3\!\!}/{\rm min}=0.3{\rm m}^{3\!\!}/{\rm min}$. Then $$\begin{eqnarray*} 0.3 &=&25(1.5) \dfrac{\textit{dh}}{\textit{dt}}\qquad\qquad \color{#0066A7}{{\hbox{\(\dfrac{dV}{dt}=25 h\dfrac{dh}{dt}\)}}} \\[3pt] \frac{\textit{dh}}{\textit{dt}} &=&\frac{0.3}{25(1.5)}=0.008 \end{eqnarray*}$$

When the height of the water is $1.5\, {\rm m}$, the water level is rising at a rate of $0.008\, {\rm m}/{\rm min}$.